What is the sum from n=1 to infinity of 2/((4n^2-1)) ?

The sum from n=1 to infinity of 2/((4n^2-1)) is 1

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sum from n=1 to infinity of 2/((4n^2-1))

n = 1 \sum \infty \frac{2}{( 4 n ^{2} - 1 )}

1

Solution steps

n = 1 \sum \infty \frac{2}{4 n ^{2} - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

2 \cdot \frac{1}{2} : 1

= 1

n = 1 \sum \infty (e)^{- n}

n = 2 \sum \infty \frac{1}{n} - \frac{1}{4 ^{n}}

n = 1 \sum \infty \frac{5 n}{5 n + 1}

k = 1 \sum \infty \frac{5}{k ( k + 2 )}

n = 1 \sum \infty \frac{3}{e ^{n} - 1}

What is the sum from n=1 to infinity of 2/((4n^2-1)) ?
The sum from n=1 to infinity of 2/((4n^2-1)) is 1